This problem is looking for the minimum value of μs that is
necessary to achieve the record time. To solve this problem:
Assuming the front wheels are off the ground for the entire
¼ mile = 402.3 m, the acceleration a = µs·9.8 m/s².
For a constant acceleration, distance = 402.3
m = 1/2at^2 = 804.6 m / (4.43 s)^2 = a = µs·9.8 m/s^2
µs = 804.6 m / (4.43s)^2 / 9.8 m/s^2 = 4.18
Answer:
a = 4.9(1 - sinθ - 0.4cosθ)
Explanation:
Really not possible without a complete setup.
I will ASSUME that this an Atwood machine with two masses (m) connected by an ideal rope passing over an ideal pulley. One mass hangs freely and the other is on a slope of angle θ to the horizontal with coefficient of friction μ. Gravity is g
F = ma
mg - mgsinθ - μmgcosθ = (m + m)a
mg(1 - sinθ - μcosθ) = 2ma
½g(1 - sinθ - μcosθ) = a
maximum acceleration is about 2.94 m/s² when θ = 0
acceleration will be zero when θ is greater than about 46.4°
Answer:
Gases are easily compressed. We can see evidence of this in Table 1 in Thermal Expansion of Solids and Liquids, where you will note that gases have the largest coefficients of volume expansion. The large coefficients mean that gases expand and contract very rapidly with temperature changes. In addition, you will note that most gases expand at the same rate, or have the same β. This raises the question as to why gases should all act in nearly the same way, when liquids and solids have widely varying expansion rates.
Explanation:
Answer:
r = 0.5 m
Explanation:
First we find the angular speed of the ball by using its period:
ω = θ/t
For the time period:
ω = angular speed = ?
θ = angular displacement = 2π rad
t = time period = 0.5 s
Therefore,
ω = 2π rad/0.5 s
ω = 12.56 rad/s
Now, for the radius:
v = rω
r = v/ω
where,
v = linear speed = 6.29 m/s
r = radius = ?
r = (6.29 m/s)/(12.56 rad/s)
<u>r = 0.5 m</u>
BMI is a measure of body fat determined by ones Height, Weight, and Gender.
